Energy Recovery from Municipal Solid Waste: Profitable Pollution Prevention at the City of Spokane, Washington (see Appendix G)

EXAMPLE 7.13 WASTE‐TO‐ENERGY

The City of Newcastle‐Upon‐Tyne, UK, incinerates some of its municipal RDF for electricity production. The city operates two 13 MW RDF boilers, which have 88% overall efficiency. The RDF has an average heat content 1650 Btu/lb. Calculate the following:

1. The daily mass of RDF (T/day) required to burn to generate the same power.
2. If coal were used in place of RDF, how much coal (in T/Y) that would be needed to burn to produce the same energy. The coal has an average heat content of 33 516 kJ/kg.

SOLUTION

This is essentially a mass and energy balance problem, in which solid wastes with significant heat content are being collected and burned to generate heat and electricity.

1. The daily mass of RDF in T/dayFirst, find input power needed.Then, mass of RDF required to produce same power
2. Mass of coal required to produce same power

EXAMPLE 7.14 WASTE‐TO‐ENERGY

Assume the proximate and ultimate analysis of a solid waste.

Proximate analysis	(wt%)	Ultimate analysis	(wt%)
Moisture	20	Moisture	20
Volatile matter	60	Carbon	30
Ash	20	Hydrogen	4
Total	100	Oxygen	25.5
		Nitrogen	0.37
		Sulfur	0.13
		Ash and metal (inerts)	20
		Total	100

Also assume the following conditions are applicable:

1. The as‐fired heating values of the solid waste are represented by the equations:(7.6)(7.7)where HHV is the high‐heating value and LHV is the low‐heating value; C, H, O, S, and W are the weight percentages of carbon, hydrogen, oxygen, sulfur, and W is the weight of the moisture, respectively. And the answer is in terms of Btu/lb.
2. The grate residue contains 10% unburned carbon.
3. The entering air is at 77 °F; the residue is at 700 °F.
4. The radiation loss is 0.006 Btu/Btu.
5. The specific heat of residue is 0.3 Btu/lb °F.
6. The latent heat of water is 1040 Btu/lb.
7. All the oxygen in the waste is bound as water.
8. The theoretical air requirements based on stoichiometry are as follows:ElementReaction with oxygenAir requirement (lb/lb)CC + O → CO11.52H2H₂ + O₂ → 2H₂O34.56SS + O₂ → SO₂4.31
9. The moisture in combustion air is negligible.
10. The net hydrogen available for combustion is equal to the fraction ().In other words, this is the percentage of hydrogen minus ⅛th the percentage oxygen. This accounts for “bound water” in the dry combustible material.
11. The heating value for carbon is 14 000 Btu/lb.Find:
    1. Estimate the HHV and LHV of this waste using the Eqs. (7.6) and (7.7).
    2. Calculate the air requirement for complete combustion of 100 T of this waste.
    3. Determine the heat available in the exhaust gases from combustion of 100 T/day waste by a heat balance, both in million Btu and kWh units.

SOLUTION

1. From Eq. (7.6),and from Eq. (7.7),
2. First, compute the weights of the elements of the waste:ElementsComputation (lb)lb/dayCarbon0.30 × 200 00060 000Hydrogen0.04 × 200 0008 000Oxygen0.255 × 200 00051 000Nitrogen0.0037 × 200 000740Sulfur0.0013 × 200 000260Ash and metals (inert)0.20 × 200 00040 000Moisture0.20 × 200 00040 000Second, compute the residue:Because the grate residue contains 10% unburnt carbon (an assumption made or this problem),Third, compute the available hydrogen and bound water:Therefore,Fourth, calculate the air required:ElementAir requirement (lb/day)Carbon = (60 000 – 4 444) × 11.52640 005Hydrogen = 1.625 × 34.5656.160Sulfur = 260 × 4.311 120Total dry theoretical air (C + H + S)697 285Total dry air (100‐percent excess)1 394 570
3. First, determine the amount of water produced by combustion of the available hydrogen.The remaining computations of heat balance for the process are as follows:ItemHeat value (10⁶ Btu/day)Gross heat input
   1065.6Heat lost in unburned carbon:
   62.21Radiation lost:
   6.39Inherent moisture:
   41.60Moisture in bound water:
   59.65Moisture from the combustion of available hydrogen:
   15.21Sensible heat in residue:
   8.30Total losses193.38Net heat available in flue gases:
   877.22Therefore, net energy available in kWh = 877.22 × 10⁶ Btu= 877.22 × 10⁶ Btu × 2.93 × 10⁻⁴ kWh/Btu = 257 000 kWhCombustion efficiency:
   81.86%